post_href stringlengths 57 213 | python_solutions stringlengths 71 22.3k | slug stringlengths 3 77 | post_title stringlengths 1 100 | user stringlengths 3 29 | upvotes int64 -20 1.2k | views int64 0 60.9k | problem_title stringlengths 3 77 | number int64 1 2.48k | acceptance float64 0.14 0.91 | difficulty stringclasses 3
values | __index_level_0__ int64 0 34k |
|---|---|---|---|---|---|---|---|---|---|---|---|
https://leetcode.com/problems/number-of-pairs-of-strings-with-concatenation-equal-to-target/discuss/1613355/Python3-one-liner | class Solution:
def numOfPairs(self, nums: List[str], target: str) -> int:
return collections.Counter(map(''.join, itertools.permutations(nums, 2))).get(target, 0) | number-of-pairs-of-strings-with-concatenation-equal-to-target | Python3 one-liner | hangyu1130 | 0 | 54 | number of pairs of strings with concatenation equal to target | 2,023 | 0.729 | Medium | 28,200 |
https://leetcode.com/problems/number-of-pairs-of-strings-with-concatenation-equal-to-target/discuss/1501678/O(n%2Bm)-T-or-O(n%2Bm)-S-or-prefix-function-%2B-hash-table-or-Python-3 | class Solution:
def numOfPairs(self, nums: List[str], target: str) -> int:
import collections
def prefix_function(string: str) -> list[int]:
"""Return prefix function for string. O(n)
:param string: input string.
:return:
"""
... | number-of-pairs-of-strings-with-concatenation-equal-to-target | O(n+m) T | O(n+m) S | prefix function + hash table | Python 3 | CiFFiRO | 0 | 58 | number of pairs of strings with concatenation equal to target | 2,023 | 0.729 | Medium | 28,201 |
https://leetcode.com/problems/number-of-pairs-of-strings-with-concatenation-equal-to-target/discuss/1499756/Python-Back-Tracking-O(N-*-2) | class Solution:
def numOfPairs(self, nums: List[str], target: str) -> int:
def backTracking(res, curr_path, start):
if len(curr_path) == 2:
res.append(curr_path)
return
for i in range(len(nums)):
if not curr_path:
b... | number-of-pairs-of-strings-with-concatenation-equal-to-target | [Python] Back Tracking O(N * 2) | Sai-Adarsh | 0 | 43 | number of pairs of strings with concatenation equal to target | 2,023 | 0.729 | Medium | 28,202 |
https://leetcode.com/problems/number-of-pairs-of-strings-with-concatenation-equal-to-target/discuss/1499214/Python-3-Solution | class Solution:
def numOfPairs(self, nums: List[str], target: str) -> int:
count=0
for i in range(0,len(nums)):
for j in range(i+1,len(nums)):
if (nums[i]+nums[j])==target:
count=count+1
if (nums[j]+nums[i])==target:
... | number-of-pairs-of-strings-with-concatenation-equal-to-target | [Python 3] Solution | evenly_odd | 0 | 48 | number of pairs of strings with concatenation equal to target | 2,023 | 0.729 | Medium | 28,203 |
https://leetcode.com/problems/number-of-pairs-of-strings-with-concatenation-equal-to-target/discuss/1499121/Python-Solution | class Solution:
def numOfPairs(self, nums: List[str], target: str) -> int:
n = len(nums)
count = 0
for i in range(n):
a = nums[i]
for j in range(i+1,n):
b = nums[j]
if a+b == target:
count += 1
if b+... | number-of-pairs-of-strings-with-concatenation-equal-to-target | [Python] Solution | SaSha59 | 0 | 45 | number of pairs of strings with concatenation equal to target | 2,023 | 0.729 | Medium | 28,204 |
https://leetcode.com/problems/number-of-pairs-of-strings-with-concatenation-equal-to-target/discuss/1700256/Python3-accepted-solution | class Solution:
def numOfPairs(self, nums: List[str], target: str) -> int:
count=0
for i in range(len(nums)):
for j in range(len(nums)):
if(i!=j):
if(nums[i] + nums[j] == target):
count+=1
return count | number-of-pairs-of-strings-with-concatenation-equal-to-target | Python3 accepted solution | sreeleetcode19 | -1 | 92 | number of pairs of strings with concatenation equal to target | 2,023 | 0.729 | Medium | 28,205 |
https://leetcode.com/problems/maximize-the-confusion-of-an-exam/discuss/1951750/WEEB-DOES-PYTHONC%2B%2B-SLIDING-WINDOW | class Solution:
def maxConsecutiveAnswers(self, string: str, k: int) -> int:
result = 0
j = 0
count1 = k
for i in range(len(string)):
if count1 == 0 and string[i] == "F":
while string[j] != "F":
j+=1
count1+=1
j+=1
if string[i] == "F":
if count1 > 0:
count1-=1
if i - j + 1... | maximize-the-confusion-of-an-exam | WEEB DOES PYTHON/C++ SLIDING WINDOW | Skywalker5423 | 2 | 95 | maximize the confusion of an exam | 2,024 | 0.598 | Medium | 28,206 |
https://leetcode.com/problems/maximize-the-confusion-of-an-exam/discuss/1624131/python-sliding-window-with-explanation | class Solution:
def maxConsecutiveAnswers(self, answerKey: str, k: int) -> int:
n = len(answerKey)
left = ret = numT = numF = 0
for right in range(n):
if answerKey[right]=='T':
numT+=1
else:
numF+=1
while numT>k and... | maximize-the-confusion-of-an-exam | python sliding window with explanation | 1579901970cg | 2 | 132 | maximize the confusion of an exam | 2,024 | 0.598 | Medium | 28,207 |
https://leetcode.com/problems/maximize-the-confusion-of-an-exam/discuss/1521886/Python-Same-Problem-as-'Longest-Consecutive-Subarray-of-1's-With-K-Deletions' | class Solution:
def maxConsecutiveAnswers(self, answerKey: str, k: int) -> int:
ifTrueIsAnswer = self.confuseStudents(answerKey, k, "T")
ifFalseIsAnswer = self.confuseStudents(answerKey, k, "F")
return max(ifTrueIsAnswer, ifFalseIsAnswer)
def confuseStudents(self, array, k, key):
... | maximize-the-confusion-of-an-exam | [Python] Same Problem as 'Longest Consecutive Subarray of 1's With K Deletions' | ramit_kumar | 2 | 124 | maximize the confusion of an exam | 2,024 | 0.598 | Medium | 28,208 |
https://leetcode.com/problems/maximize-the-confusion-of-an-exam/discuss/1567232/Python-solution-212-ms-better-than-98 | class Solution:
def maxConsecutiveAnswers(self, answerKey: str, k: int) -> int:
tmpK = k
#check for False
i = 0
for j in range(0,len(answerKey)):
if answerKey[j] == 'T':
k -= 1
if k < 0:
if answerKey[i] == 'T':
... | maximize-the-confusion-of-an-exam | Python solution 212 ms better than 98% | akshaykumar19002 | 1 | 100 | maximize the confusion of an exam | 2,024 | 0.598 | Medium | 28,209 |
https://leetcode.com/problems/maximize-the-confusion-of-an-exam/discuss/1499015/Python3-sliding-window | class Solution:
def maxConsecutiveAnswers(self, answerKey: str, k: int) -> int:
def fn(target):
"""Return max consecutive target."""
ans = cnt = ii = 0
for i, x in enumerate(answerKey):
if x == target: cnt += 1
while cnt > k:
... | maximize-the-confusion-of-an-exam | [Python3] sliding window | ye15 | 1 | 60 | maximize the confusion of an exam | 2,024 | 0.598 | Medium | 28,210 |
https://leetcode.com/problems/maximize-the-confusion-of-an-exam/discuss/2676084/Ezy-to-understand-two-pointers-python3-solution-or-O(n)-time-or-O(1)-space | class Solution:
# O(n) time,
# O(1) space,
# Approach: two pointers,
def maxConsecutiveAnswers(self, answerKey: str, k: int) -> int:
n = len(answerKey)
k_copy = k
longst_true = 0
l, r = 0, 0
while r < n:
if answerKey[r] == 'T':
... | maximize-the-confusion-of-an-exam | Ezy to understand two pointers python3 solution | O(n) time | O(1) space | destifo | 0 | 9 | maximize the confusion of an exam | 2,024 | 0.598 | Medium | 28,211 |
https://leetcode.com/problems/maximize-the-confusion-of-an-exam/discuss/2646267/Python3-Solution-oror-O(N)-Time-and-O(1)-Space-Complexity | class Solution:
def maxConsecutiveAnswers(self, answerKey: str, k: int) -> int:
countTrue=0
countFalse=0
maxTF=0
start=0
n=len(answerKey)
for i in range(n):
if answerKey[i]=="T":
countTrue+=1
else:
countFalse+=1
... | maximize-the-confusion-of-an-exam | Python3 Solution || O(N) Time & O(1) Space Complexity | akshatkhanna37 | 0 | 8 | maximize the confusion of an exam | 2,024 | 0.598 | Medium | 28,212 |
https://leetcode.com/problems/maximize-the-confusion-of-an-exam/discuss/2310680/Python3-or-Sliding-Window-Technique | class Solution:
def maxConsecutiveAnswers(self, answerKey: str, k: int) -> int:
#Time-Complexity: O(n)
#Space-Complexity: O(1)
#sliding window technique!
L = 0
hashmap = {'T': 0,
'F': 0}
answer = 0
for R in range(l... | maximize-the-confusion-of-an-exam | Python3 | Sliding Window Technique | JOON1234 | 0 | 29 | maximize the confusion of an exam | 2,024 | 0.598 | Medium | 28,213 |
https://leetcode.com/problems/maximize-the-confusion-of-an-exam/discuss/2175999/Using-the-concept-and-approach-of-max-consecutive-ones-iii-oror-Sliding-Window-oror-Simplest-Solution | class Solution:
def longestConsecutive(self, array, k):
length = len(array)
l = 0
for r, val in enumerate(array):
k -= (1 - val)
if k < 0:
k += (1 - array[l])
l += 1
return r - l + 1
def maxConsecutiveAnswers(self,... | maximize-the-confusion-of-an-exam | Using the concept and approach of max consecutive ones iii || Sliding Window || Simplest Solution | Vaibhav7860 | 0 | 26 | maximize the confusion of an exam | 2,024 | 0.598 | Medium | 28,214 |
https://leetcode.com/problems/maximize-the-confusion-of-an-exam/discuss/2010539/Python-160ms | class Solution:
def maxConsecutiveAnswers(self, answerKey: str, k: int) -> int:
t = f = mx = i = 0
for j in range(len(answerKey)):
if answerKey[j] == 'T':
t += 1
else:
f += 1
if t > k and f > k: # invalid - just move 1 step
... | maximize-the-confusion-of-an-exam | Python 160ms | qeisar | 0 | 42 | maximize the confusion of an exam | 2,024 | 0.598 | Medium | 28,215 |
https://leetcode.com/problems/maximize-the-confusion-of-an-exam/discuss/1928657/6-Lines-Python-Solution-oror-90-Faster-oror-Memory-less-than-88 | class Solution:
def maxConsecutiveAnswers(self, A: str, k: int) -> int:
w=0 ; i=0 ; C=Counter()
for j in range(len(A)):
C[A[j]]+=1
if C['T']>k and C['F']>k: C[A[i]]-=1 ; i+=1
else: w=max(w,j-i+1)
return w | maximize-the-confusion-of-an-exam | 6-Lines Python Solution || 90% Faster || Memory less than 88% | Taha-C | 0 | 62 | maximize the confusion of an exam | 2,024 | 0.598 | Medium | 28,216 |
https://leetcode.com/problems/maximize-the-confusion-of-an-exam/discuss/1827576/Python-easy-to-read-and-understand-or-Leetcode-424 | class Solution:
def maxConsecutiveAnswers(self, answerKey: str, k: int) -> int:
s = answerKey[:]
d = {}
ans, i = 0, 0
for j in range(len(s)):
d[s[j]] = d.get(s[j], 0) + 1
cnt = (j - i + 1) - max(d.values())
while cnt > k:
d[s[i]] -=... | maximize-the-confusion-of-an-exam | Python easy to read and understand | Leetcode 424 | sanial2001 | 0 | 97 | maximize the confusion of an exam | 2,024 | 0.598 | Medium | 28,217 |
https://leetcode.com/problems/maximize-the-confusion-of-an-exam/discuss/1743620/Python3-solution-using-two-helper-functions | class Solution:
def maxConsecutiveAnswers(self, answerkey: str, k: int) -> int:
from collections import defaultdict
mydict1=defaultdict(int)
mydict2=defaultdict(int)
n=len(answerkey)
def helper1(k):
left=0
ans=0
for right ... | maximize-the-confusion-of-an-exam | Python3 solution using two helper functions | Karna61814 | 0 | 31 | maximize the confusion of an exam | 2,024 | 0.598 | Medium | 28,218 |
https://leetcode.com/problems/maximize-the-confusion-of-an-exam/discuss/1544711/python3-2-Pass-slide-window-with-Queue | class Solution:
def maxConsecutiveAnswers(self, answerKey: str, k: int) -> int:
# idea is based on 2 pass solution with sliding window
# one pass is for changing T by F and count maximum subStr len of all Fs
# other pass is for changing F by T and count maximum subStr len of all Ts
... | maximize-the-confusion-of-an-exam | python3 , 2 Pass , slide window with Queue | Code-IQ7 | 0 | 28 | maximize the confusion of an exam | 2,024 | 0.598 | Medium | 28,219 |
https://leetcode.com/problems/maximize-the-confusion-of-an-exam/discuss/1499152/Python-Sliding-Window | class Solution:
def maxConsecutiveAnswers(self, nums: str, k: int) -> int:
n = len(nums)
def util(char, K):
ans, l = 0, 0
for r in range(n):
if nums[r] == char:
if K == 0:
while nums[l] != char:
... | maximize-the-confusion-of-an-exam | Python Sliding Window | abkc1221 | 0 | 56 | maximize the confusion of an exam | 2,024 | 0.598 | Medium | 28,220 |
https://leetcode.com/problems/maximum-number-of-ways-to-partition-an-array/discuss/1499024/Python3-binary-search | class Solution:
def waysToPartition(self, nums: List[int], k: int) -> int:
prefix = [0]
loc = defaultdict(list)
for i, x in enumerate(nums):
prefix.append(prefix[-1] + x)
if i < len(nums)-1: loc[prefix[-1]].append(i)
ans = 0
if prefix[-1] % ... | maximum-number-of-ways-to-partition-an-array | [Python3] binary search | ye15 | 2 | 227 | maximum number of ways to partition an array | 2,025 | 0.321 | Hard | 28,221 |
https://leetcode.com/problems/maximum-number-of-ways-to-partition-an-array/discuss/2801817/Python3-Counter-of-(Left-Right)-at-Each-Pivot | class Solution:
def waysToPartition(self, nums: List[int], k: int) -> int:
"""This is a good problem. It's not difficult, but is quite complex.
The idea is that once we change a position at i, for all the pivots at
1...i, the sum of the left half stay the same whereas the sum of
the... | maximum-number-of-ways-to-partition-an-array | [Python3] Counter of (Left - Right) at Each Pivot | FanchenBao | 0 | 2 | maximum number of ways to partition an array | 2,025 | 0.321 | Hard | 28,222 |
https://leetcode.com/problems/minimum-moves-to-convert-string/discuss/1500215/Python3-scan | class Solution:
def minimumMoves(self, s: str) -> int:
ans = i = 0
while i < len(s):
if s[i] == "X":
ans += 1
i += 3
else: i += 1
return ans | minimum-moves-to-convert-string | [Python3] scan | ye15 | 28 | 1,200 | minimum moves to convert string | 2,027 | 0.537 | Easy | 28,223 |
https://leetcode.com/problems/minimum-moves-to-convert-string/discuss/1525838/Easy-Python-Solution-or-Faster-than-99-(24-ms) | class Solution:
def minimumMoves(self, s: str) -> int:
i, m = 0, 0
l = len(s)
while i < l:
if s[i] != 'X':
i += 1
elif 'X' not in s[i:i+1]:
i += 2
elif 'X' in s[i:i+2]:
m += 1
i += 3
... | minimum-moves-to-convert-string | Easy Python Solution | Faster than 99% (24 ms) | the_sky_high | 5 | 351 | minimum moves to convert string | 2,027 | 0.537 | Easy | 28,224 |
https://leetcode.com/problems/minimum-moves-to-convert-string/discuss/1722144/Simplest-Python-3-code | class Solution:
def minimumMoves(self, s: str) -> int:
sl=list(s)
out=0
for i in range(0,len(sl)-2):
if sl[i]=="X":
sl[i]="O"
sl[i+1]="O"
sl[i+2]="O"
out+=1
elif sl[i]=="O":
continue
... | minimum-moves-to-convert-string | Simplest Python 3 code | rajatkumarrrr | 1 | 92 | minimum moves to convert string | 2,027 | 0.537 | Easy | 28,225 |
https://leetcode.com/problems/minimum-moves-to-convert-string/discuss/1512345/Python-simple-solution-better-than-97 | class Solution(object):
def minimumMoves(self, s):
"""
:type s: str
:rtype: int
"""
ans, l = 0, 0
while l < len(s):
if s[l] == 'X':
l+=3
ans +=1
else:
l+=1
return ans | minimum-moves-to-convert-string | Python simple solution better than 97% | avigupta10 | 1 | 116 | minimum moves to convert string | 2,027 | 0.537 | Easy | 28,226 |
https://leetcode.com/problems/minimum-moves-to-convert-string/discuss/2777309/Python-intiutive-O(N)-time-and-O(1)-solution | class Solution:
def minimumMoves(self, s: str) -> int:
i = 0
count = 0
while i<len(s):
if s[i]== "X":
count+=1
i +=3
else:
i+=1
return count | minimum-moves-to-convert-string | Python intiutive O(N) time and O(1) solution | Rajeev_varma008 | 0 | 2 | minimum moves to convert string | 2,027 | 0.537 | Easy | 28,227 |
https://leetcode.com/problems/minimum-moves-to-convert-string/discuss/2640186/Python3-Simple-Iteration | class Solution:
def minimumMoves(self, s: str) -> int:
x = list(s)
x.append('0')
x.append('0')
c = 0
for i in range(len(x) - 2):
if x[i] == 'X':
c+=1
x[i+1] = '0'
x[i+2] = '0'
return c | minimum-moves-to-convert-string | Python3 Simple Iteration | godshiva | 0 | 5 | minimum moves to convert string | 2,027 | 0.537 | Easy | 28,228 |
https://leetcode.com/problems/minimum-moves-to-convert-string/discuss/2551670/simple-python-solution | class Solution:
def minimumMoves(self, s: str) -> int:
x, y = 0,0
while x < len(s):
if s[x] == 'X':
x += 3
y += 1
else:
x += 1
return y | minimum-moves-to-convert-string | simple python solution | maschwartz5006 | 0 | 28 | minimum moves to convert string | 2,027 | 0.537 | Easy | 28,229 |
https://leetcode.com/problems/minimum-moves-to-convert-string/discuss/2506930/Single-pointer | class Solution:
def minimumMoves(self, s: str) -> int:
# use a single pointer to locate element at index
# if the element is X increment result and move the pointer by 3 (s[i:i+3] = "O")
# if not, move the pointer by 1
# Time O(N) Space: O(1)
n = len(s)
res =... | minimum-moves-to-convert-string | Single pointer | andrewnerdimo | 0 | 14 | minimum moves to convert string | 2,027 | 0.537 | Easy | 28,230 |
https://leetcode.com/problems/minimum-moves-to-convert-string/discuss/2115342/Python-3%3A-Easy-to-understand-(Faster-than-99)-oror-Self-Explanatory | class Solution:
def minimumMoves(self, s: str) -> int:
output=0
while "X" in s:
i=s.find('X')
output+=1
s=s[:i]+s[i+3:]
return output | minimum-moves-to-convert-string | Python 3: Easy to understand (Faster than 99%) || Self-Explanatory | kushal2201 | 0 | 85 | minimum moves to convert string | 2,027 | 0.537 | Easy | 28,231 |
https://leetcode.com/problems/minimum-moves-to-convert-string/discuss/1999896/python-3-oror-greedy-solution-oror-O(n)O(1) | class Solution:
def minimumMoves(self, s: str) -> int:
i = res = 0
n = len(s)
while i < n:
if s[i] == 'X':
res += 1
i += 3
else:
i += 1
return res | minimum-moves-to-convert-string | python 3 || greedy solution || O(n)/O(1) | dereky4 | 0 | 90 | minimum moves to convert string | 2,027 | 0.537 | Easy | 28,232 |
https://leetcode.com/problems/minimum-moves-to-convert-string/discuss/1881593/python-dollarolution | class Solution:
def minimumMoves(self, s: str) -> int:
i = 0
count = 0
while i < len(s):
if s[i] == 'X':
count += 1
i += 3
continue
i += 1
return count | minimum-moves-to-convert-string | python $olution | AakRay | 0 | 45 | minimum moves to convert string | 2,027 | 0.537 | Easy | 28,233 |
https://leetcode.com/problems/minimum-moves-to-convert-string/discuss/1795983/Simple-Python-Solution-oror-99-Faster-(20ms)-oror-Memory-less-than-99 | class Solution:
def minimumMoves(self, s: str) -> int:
ans, i = 0, 0
while i< len(s):
if s[i]=='O':
i += 1
continue
else:
ans += 1
i += 3
return ans | minimum-moves-to-convert-string | Simple Python Solution || 99% Faster (20ms) || Memory less than 99% | Taha-C | 0 | 59 | minimum moves to convert string | 2,027 | 0.537 | Easy | 28,234 |
https://leetcode.com/problems/minimum-moves-to-convert-string/discuss/1642870/Python-O(n)-time-O(1)-space-easy-solution | class Solution:
def minimumMoves(self, s: str) -> int:
n = len(s)
idx = 0
res = 0
while idx < n:
if s[idx] == 'O':
idx += 1
else:
res += 1
idx += 3
return res | minimum-moves-to-convert-string | Python O(n) time, O(1) space easy solution | byuns9334 | 0 | 69 | minimum moves to convert string | 2,027 | 0.537 | Easy | 28,235 |
https://leetcode.com/problems/minimum-moves-to-convert-string/discuss/1508251/Greedy-O(n)-solution-in-Python | class Solution:
def minimumMoves(self, s: str) -> int:
i = cnt = 0
while i < len(s):
if s[i] == "X":
cnt += 1
i += 3
else:
i += 1
return cnt | minimum-moves-to-convert-string | Greedy O(n) solution in Python | mousun224 | 0 | 62 | minimum moves to convert string | 2,027 | 0.537 | Easy | 28,236 |
https://leetcode.com/problems/minimum-moves-to-convert-string/discuss/1501060/Python-one-pass-O(N) | class Solution:
def minimumMoves(self, s: str) -> int:
idx = moves = 0
while idx < len(s):
if s[idx] == 'X':
moves += 1
idx += 3
else:
idx += 1
return moves | minimum-moves-to-convert-string | Python, one pass O(N) | blue_sky5 | 0 | 49 | minimum moves to convert string | 2,027 | 0.537 | Easy | 28,237 |
https://leetcode.com/problems/minimum-moves-to-convert-string/discuss/1500345/Python-or-O(n)-time-O(1)-Space-or-6-lines | class Solution:
def minimumMoves(self, s: str) -> int:
min_moves = i = 0
while i < len(s):
is_x = s[i] == 'X'
min_moves += is_x
i += 2*is_x + 1
return min_moves | minimum-moves-to-convert-string | Python | O(n) time O(1) Space | 6 lines | leeteatsleep | 0 | 52 | minimum moves to convert string | 2,027 | 0.537 | Easy | 28,238 |
https://leetcode.com/problems/minimum-moves-to-convert-string/discuss/1700019/Python3-accepted-solution | class Solution:
def minimumMoves(self, s: str) -> int:
count=0
li = []
for i in range(0,len(s)-2):
if(s[i]=="X"):
count+=1
s = s[:i] + "OOO" + s[i+3:]
else:
continue
if(s[-3:]=="OXX" or s[-3:]=="OOX" or s[-3:]=="... | minimum-moves-to-convert-string | Python3 accepted solution | sreeleetcode19 | -1 | 58 | minimum moves to convert string | 2,027 | 0.537 | Easy | 28,239 |
https://leetcode.com/problems/minimum-moves-to-convert-string/discuss/1501346/Python3-or-Greedy-Intuitive | class Solution:
def safe_val_by_index(self, arr, start, end):
try:
val = arr[start:end]
except IndexError:
return []
return val
def minimumMoves(self, s: str) -> int:
a = list(s)
n = len(a)
p = 'X'
template =... | minimum-moves-to-convert-string | [Python3] | Greedy / Intuitive | patefon | -1 | 40 | minimum moves to convert string | 2,027 | 0.537 | Easy | 28,240 |
https://leetcode.com/problems/find-missing-observations/discuss/1506196/Divmod-and-list-comprehension-96-speed | class Solution:
def missingRolls(self, rolls: List[int], mean: int, n: int) -> List[int]:
missing_val, rem = divmod(mean * (len(rolls) + n) - sum(rolls), n)
if rem == 0:
if 1 <= missing_val <= 6:
return [missing_val] * n
elif 1 <= missing_val < 6:
retu... | find-missing-observations | Divmod and list comprehension, 96% speed | EvgenySH | 2 | 133 | find missing observations | 2,028 | 0.439 | Medium | 28,241 |
https://leetcode.com/problems/find-missing-observations/discuss/1500222/Python3-greedy | class Solution:
def missingRolls(self, rolls: List[int], mean: int, n: int) -> List[int]:
total = mean * (len(rolls) + n) - sum(rolls)
if not n <= total <= 6*n: return []
q, r = divmod(total, n)
return [q]*(n-r) + [q+1]*r | find-missing-observations | [Python3] greedy | ye15 | 1 | 73 | find missing observations | 2,028 | 0.439 | Medium | 28,242 |
https://leetcode.com/problems/find-missing-observations/discuss/2820752/Python-No-loops-solutions | class Solution:
def missingRolls(self, rolls, mean, n):
m = len(rolls)
sum_target = mean * (n + m)
sum_current = sum(rolls)
sum_remaining = sum_target - sum_current
if not (n <= sum_remaining <= 6*n):
return []
remaining_rolls = [sum_remaining // ... | find-missing-observations | [Python] No loops solutions | bison_a_besoncon | 0 | 3 | find missing observations | 2,028 | 0.439 | Medium | 28,243 |
https://leetcode.com/problems/find-missing-observations/discuss/2669271/Python-or-Simple-Maths | class Solution:
def missingRolls(self, rolls: List[int], mean: int, n: int) -> List[int]:
m = len(rolls)
sum_rolls = sum(rolls)
target = mean * (m + n) - sum_rolls
q = target // n
r = target % n
if q > 6 or q < 1 or (q == 6 and r > 0):
return []
... | find-missing-observations | Python | Simple Maths | on_danse_encore_on_rit_encore | 0 | 8 | find missing observations | 2,028 | 0.439 | Medium | 28,244 |
https://leetcode.com/problems/find-missing-observations/discuss/1694949/python-simple-solution | class Solution:
def missingRolls(self, rolls: List[int], mean: int, n: int) -> List[int]:
m, s = len(rolls), sum(rolls)
x = mean*(m+n)-s
if x < n or x > 6*n:
return []
else:
t = x//n
res = [t for _ in range(n-1)]
res.append(... | find-missing-observations | python simple solution | byuns9334 | 0 | 96 | find missing observations | 2,028 | 0.439 | Medium | 28,245 |
https://leetcode.com/problems/find-missing-observations/discuss/1686301/WEEB-DOES-PYTHON-USING-MATH | class Solution:
def missingRolls(self, rolls: List[int], mean: int, n: int) -> List[int]:
numTerms = n + len(rolls)
sumOfM = sum(rolls)
sumOfN = mean * numTerms - sumOfM
# (die cannot be > 6) or (sumOfN cannot be negative or die cannot be less than 0)
if sumOfN / n > 6 or sumOfN // n <= 0:
return []
... | find-missing-observations | WEEB DOES PYTHON USING MATH | Skywalker5423 | 0 | 52 | find missing observations | 2,028 | 0.439 | Medium | 28,246 |
https://leetcode.com/problems/find-missing-observations/discuss/1502434/100-Faster-python-solution | class Solution:
def missingRolls(self, rolls: List[int], mean: int, n: int) -> List[int]:
m = len(rolls)
# Mean = sum(rolls) + n*average_n /m+n :-> average = n*average_n
average = mean*(m+n) - sum(rolls)
# Cheak: average cannot be greater than n*6 beacuse maximum value of dice is 6
# ... | find-missing-observations | 100% Faster python solution | ce17b127 | 0 | 45 | find missing observations | 2,028 | 0.439 | Medium | 28,247 |
https://leetcode.com/problems/find-missing-observations/discuss/1500378/Easy-Approach-oror-Math-oror-well-Explained | class Solution:
def missingRolls(self, rolls: List[int], mean: int, n: int) -> List[int]:
m = len(rolls)
req = mean*(m+n) - sum(rolls)
q,r = divmod(req,n)
if q<=0 or req>n*6:
return []
res = [q]*n
for i in range(r):
res[i]+=1
return res | find-missing-observations | 📌📌 Easy-Approach || Math || well-Explained 🐍 | abhi9Rai | 0 | 41 | find missing observations | 2,028 | 0.439 | Medium | 28,248 |
https://leetcode.com/problems/find-missing-observations/discuss/1500161/Python-O(n)-solution | class Solution:
def missingRolls(self, rolls: List[int], mean: int, n: int) -> List[int]:
#finding the equal distribution of remaining sum
temp = mean*(n+len(rolls)) - sum(rolls)
each = temp//n
rem = temp%n
#return empty if out of dice range
if (not (1<=(each... | find-missing-observations | Python O(n) solution | abkc1221 | 0 | 160 | find missing observations | 2,028 | 0.439 | Medium | 28,249 |
https://leetcode.com/problems/stone-game-ix/discuss/1500343/Python3-freq-table | class Solution:
def stoneGameIX(self, stones: List[int]) -> bool:
freq = defaultdict(int)
for x in stones: freq[x % 3] += 1
if freq[0]%2 == 0: return freq[1] and freq[2]
return abs(freq[1] - freq[2]) >= 3 | stone-game-ix | [Python3] freq table | ye15 | 3 | 165 | stone game ix | 2,029 | 0.264 | Medium | 28,250 |
https://leetcode.com/problems/stone-game-ix/discuss/1501697/python-clean-and-short-solution | class Solution:
def stoneGameIX(self, stones: List[int]) -> bool:
stones = [v % 3 for v in stones]
d = defaultdict(int)
for v in stones:
d[v] += 1
while d[1] >= 2 and d[2] >= 2:
d[2] -= 1
d[1] -= 1
if d[0] % 2 == ... | stone-game-ix | python clean and short solution | pureme | 1 | 107 | stone game ix | 2,029 | 0.264 | Medium | 28,251 |
https://leetcode.com/problems/smallest-k-length-subsequence-with-occurrences-of-a-letter/discuss/1502134/PYTHON3-O(n)-using-stack-with-explanation | class Solution:
def smallestSubsequence(self, s: str, k: int, letter: str, repetition: int) -> str:
counts,total = 0, 0
n = len(s)
for ch in s:
if ch==letter:
total +=1
stack = []
occ = 0
for idx,ch in enumerate(s):
if ch==lette... | smallest-k-length-subsequence-with-occurrences-of-a-letter | [PYTHON3] O(n) using stack with explanation | irt | 2 | 300 | smallest k length subsequence with occurrences of a letter | 2,030 | 0.387 | Hard | 28,252 |
https://leetcode.com/problems/smallest-k-length-subsequence-with-occurrences-of-a-letter/discuss/1860026/Python-Stack-solution | class Solution:
def smallestSubsequence(self, s: str, k: int, letter: str, repetition: int) -> str:
s = list(s)
stack = []
countAll = s.count(letter)
count = 0
for ind, i in enumerate(s):
while stack and stack[-1] > i:
if stack[-1] == letter and i ... | smallest-k-length-subsequence-with-occurrences-of-a-letter | Python Stack solution | khanter | 1 | 171 | smallest k length subsequence with occurrences of a letter | 2,030 | 0.387 | Hard | 28,253 |
https://leetcode.com/problems/smallest-k-length-subsequence-with-occurrences-of-a-letter/discuss/1500303/Python3-10-line-stack | class Solution:
def smallestSubsequence(self, s: str, k: int, letter: str, repetition: int) -> str:
rest = sum(x == letter for x in s)
stack = []
for i, x in enumerate(s):
while stack and stack[-1] > x and len(stack) + len(s) - i > k and (stack[-1] != letter or repetition < rest... | smallest-k-length-subsequence-with-occurrences-of-a-letter | [Python3] 10-line stack | ye15 | 0 | 203 | smallest k length subsequence with occurrences of a letter | 2,030 | 0.387 | Hard | 28,254 |
https://leetcode.com/problems/two-out-of-three/discuss/1513311/Python3-set | class Solution:
def twoOutOfThree(self, nums1: List[int], nums2: List[int], nums3: List[int]) -> List[int]:
s1, s2, s3 = set(nums1), set(nums2), set(nums3)
return (s1&s2) | (s2&s3) | (s1&s3) | two-out-of-three | [Python3] set | ye15 | 20 | 1,400 | two out of three | 2,032 | 0.726 | Easy | 28,255 |
https://leetcode.com/problems/two-out-of-three/discuss/1513311/Python3-set | class Solution:
def twoOutOfThree(self, nums1: List[int], nums2: List[int], nums3: List[int]) -> List[int]:
freq = Counter()
for nums in nums1, nums2, nums3: freq.update(set(nums))
return [k for k, v in freq.items() if v >= 2] | two-out-of-three | [Python3] set | ye15 | 20 | 1,400 | two out of three | 2,032 | 0.726 | Easy | 28,256 |
https://leetcode.com/problems/two-out-of-three/discuss/1525692/Easy-Python-Solution-or-Faster-than-97-(64-ms) | class Solution:
def twoOutOfThree(self, nums1: List[int], nums2: List[int], nums3: List[int]) -> List[int]:
ret = []
ret += set(nums1).intersection(set(nums2))
ret += set(nums1).intersection(set(nums3))
ret += set(nums2).intersection(set(nums3))
return set(ret) | two-out-of-three | Easy Python Solution | Faster than 97% (64 ms) | the_sky_high | 7 | 660 | two out of three | 2,032 | 0.726 | Easy | 28,257 |
https://leetcode.com/problems/two-out-of-three/discuss/1553645/python3-easy-set-solution | class Solution:
def twoOutOfThree(self, nums1: List[int], nums2: List[int], nums3: List[int]) -> List[int]:
set1 = set(nums1)
set2 = set(nums2)
set3 = set(nums3)
set12 = set1.intersection(set2)
set23 = set2.intersection(set3)
set13 = set1.intersection(set3)
... | two-out-of-three | python3, easy set solution | sirenescx | 3 | 263 | two out of three | 2,032 | 0.726 | Easy | 28,258 |
https://leetcode.com/problems/two-out-of-three/discuss/2138556/PYTHON-or-Simple-python-solution | class Solution:
def twoOutOfThree(self, nums1: List[int], nums2: List[int], nums3: List[int]) -> List[int]:
countMap = {}
for i in set(nums1):
countMap[i] = 1 + countMap.get(i, 0)
for i in set(nums2):
countMap[i] = 1 + countMap.get(i, 0)
... | two-out-of-three | PYTHON | Simple python solution | shreeruparel | 2 | 152 | two out of three | 2,032 | 0.726 | Easy | 28,259 |
https://leetcode.com/problems/two-out-of-three/discuss/1859650/Python-efficient-solution | class Solution:
def twoOutOfThree(self, nums1: List[int], nums2: List[int], nums3: List[int]) -> List[int]:
all_distinct = set(nums1 + nums2 + nums3)
count = 0
res = []
for i in all_distinct:
if i in nums1:
count += 1
if i in nums2:
... | two-out-of-three | Python efficient solution | alishak1999 | 1 | 104 | two out of three | 2,032 | 0.726 | Easy | 28,260 |
https://leetcode.com/problems/two-out-of-three/discuss/1566419/Python3-solution-faster-than-99.04-Solutions | class Solution:
def twoOutOfThree(self, nums1: List[int], nums2: List[int], nums3: List[int]) -> List[int]:
arr = []
arr.extend(list(set(nums1)))
arr.extend(list(set(nums2)))
arr.extend(list(set(nums3)))
hm = {}
for num in arr:
i... | two-out-of-three | Python3 solution faster than 99.04% Solutions | risabhmishra19 | 1 | 136 | two out of three | 2,032 | 0.726 | Easy | 28,261 |
https://leetcode.com/problems/two-out-of-three/discuss/1514549/Python3-One-line | class Solution:
def twoOutOfThree(self, nums1: List[int], nums2: List[int], nums3: List[int]) -> List[int]:
return list(set(nums1) & set(nums2) | set(nums1) & set(nums3) | set(nums2) & set(nums3)) | two-out-of-three | Python3 One line | frolovdmn | 1 | 92 | two out of three | 2,032 | 0.726 | Easy | 28,262 |
https://leetcode.com/problems/two-out-of-three/discuss/2812606/easy | class Solution:
def twoOutOfThree(self, nums1: List[int], nums2: List[int], nums3: List[int]) -> List[int]:
nums1 = set(nums1)
nums2 = set(nums2)
nums3 = set(nums3)
return list((nums1 & nums2) | (nums2 & nums3) | (nums3 & nums1)) | two-out-of-three | easy | nishithakonuganti | 0 | 2 | two out of three | 2,032 | 0.726 | Easy | 28,263 |
https://leetcode.com/problems/two-out-of-three/discuss/2800292/python-or-easy-to-understand-or-default-dict | class Solution:
def twoOutOfThree(self, nums1: List[int], nums2: List[int], nums3: List[int]) -> List[int]:
h = collections.defaultdict(int)
for item in set(nums1):
h[item] += 1
for item in set(nums2):
h[item] += 1
for item in set(nums3):
... | two-out-of-three | python | easy to understand | default dict | IAMdkk | 0 | 3 | two out of three | 2,032 | 0.726 | Easy | 28,264 |
https://leetcode.com/problems/two-out-of-three/discuss/2781234/Python-set-intersection-and-union | class Solution:
def twoOutOfThree(self, nums1: List[int], nums2: List[int], nums3: List[int]) -> List[int]:
nums1, nums2, nums3 = set(nums1), set(nums2), set(nums3)
answer = list(nums1.intersection(nums2).union(nums2.intersection(nums3)).union(nums1.intersection(nums3)))
return a... | two-out-of-three | Python set, intersection and union | Osama_Qutait | 0 | 5 | two out of three | 2,032 | 0.726 | Easy | 28,265 |
https://leetcode.com/problems/two-out-of-three/discuss/2736096/python-one-line-solution-99.63-faster | class Solution:
def twoOutOfThree(self, nums1: List[int], nums2: List[int], nums3: List[int]) -> List[int]:
return list(set(nums1)&set(nums2) | set(nums1)&set(nums3) | set(nums2)&set(nums3)) | two-out-of-three | python one line solution 99.63% faster | arifkhan1990 | 0 | 13 | two out of three | 2,032 | 0.726 | Easy | 28,266 |
https://leetcode.com/problems/two-out-of-three/discuss/2730794/Very-simple-and-intuitive-solution-using-Python-sets | class Solution:
def twoOutOfThree(self, nums1: List[int], nums2: List[int], nums3: List[int]) -> List[int]:
s1 = set(nums1)
s2 = set(nums2)
s3 = set(nums3)
i12 = s1.intersection(s2)
i23 = s2.intersection(s3)
i13 = s1.intersection(s3)
return list(i12.union(i2... | two-out-of-three | Very simple and intuitive solution using Python sets | thematrixmaster | 0 | 3 | two out of three | 2,032 | 0.726 | Easy | 28,267 |
https://leetcode.com/problems/two-out-of-three/discuss/2707042/Using-set() | class Solution:
def twoOutOfThree(self, nums1: List[int], nums2: List[int], nums3: List[int]) -> List[int]:
s1,s2,s3 = set(nums1),set(nums2),set(nums3)
return (s1&s2) | (s2&s3) | (s1&s3) | two-out-of-three | Using set() | sanjeevpathak | 0 | 3 | two out of three | 2,032 | 0.726 | Easy | 28,268 |
https://leetcode.com/problems/two-out-of-three/discuss/2705896/Easy-using-counters-and-set | class Solution:
def twoOutOfThree(self, nums1: List[int], nums2: List[int], nums3: List[int]) -> List[int]:
nums1=list(set(nums1))
nums2=list(set(nums2))
nums3=list(set(nums3))
d=nums1+nums2+nums3
dc=Counter(d)
res=[]
for i,j in dc.items():
if(j>1)... | two-out-of-three | Easy using counters and set | Raghunath_Reddy | 0 | 8 | two out of three | 2,032 | 0.726 | Easy | 28,269 |
https://leetcode.com/problems/two-out-of-three/discuss/2702297/Python-solution-clean-code-with-full-comments.-95.68-speed | class Solution:
def twoOutOfThree(self, nums1: List[int], nums2: List[int], nums3: List[int]) -> List[int]:
a = list_to_set(nums1)
b = list_to_set(nums2)
c = list_to_set(nums3)
return compare_two_sets(a, b) | compare_two_sets(a, c) | compare_two_sets(b, c)
... | two-out-of-three | Python solution, clean code with full comments. 95.68% speed | 375d | 0 | 24 | two out of three | 2,032 | 0.726 | Easy | 28,270 |
https://leetcode.com/problems/two-out-of-three/discuss/2643902/100-EASY-TO-UNDERSTANDSIMPLECLEAN | class Solution:
def twoOutOfThree(self, nums1: List[int], nums2: List[int], nums3: List[int]) -> List[int]:
output = []
for i in nums1:
if i in nums2 or i in nums3:
if i not in output:
output.append(i)
for j in nums2:
if j in nums3 ... | two-out-of-three | 🔥100% EASY TO UNDERSTAND/SIMPLE/CLEAN🔥 | YuviGill | 0 | 51 | two out of three | 2,032 | 0.726 | Easy | 28,271 |
https://leetcode.com/problems/two-out-of-three/discuss/2640209/Python3-One-Liner | class Solution:
def twoOutOfThree(self, nums1: List[int], nums2: List[int], nums3: List[int]) -> List[int]:
return [k for k, c in Counter(list(set(nums1))+list(set(nums2))+list(set(nums3))).items() if c>1] | two-out-of-three | Python3 One Liner | godshiva | 0 | 3 | two out of three | 2,032 | 0.726 | Easy | 28,272 |
https://leetcode.com/problems/two-out-of-three/discuss/2581512/Python-85-Faster-East-to-Understand-Breakdown | class Solution:
def twoOutOfThree(self, nums1: List[int], nums2: List[int], nums3: List[int]) -> List[int]:
#distinct list to iterate
dlist = list(set(nums1+nums2+nums3))
#concate list to check, distinct sublist
clist = list(set(nums1))+list(set(nums2))+list(set(num... | two-out-of-three | Python 85% Faster - East to Understand Breakdown | ovidaure | 0 | 47 | two out of three | 2,032 | 0.726 | Easy | 28,273 |
https://leetcode.com/problems/two-out-of-three/discuss/2545039/Python-Scalable-Solution-(without-Sets) | class Solution:
def twoOutOfThree(self, nums1: List[int], nums2: List[int], nums3: List[int], minCount=2) -> List[int]:
# make a list of array (so this function also works for n number of arrays)
nums_list = [nums1, nums2, nums3]
# make an array to count the values
... | two-out-of-three | [Python] - Scalable Solution (without Sets) | Lucew | 0 | 18 | two out of three | 2,032 | 0.726 | Easy | 28,274 |
https://leetcode.com/problems/two-out-of-three/discuss/2542544/Python-fast-solution-using-set-intersection-and-union | class Solution:
def twoOutOfThree(self, nums1: List[int], nums2: List[int], nums3: List[int]) -> List[int]:
set_nums1 = set(nums1)
set_nums2 = set(nums2)
set_nums3 = set(nums3)
intersect1 = set_nums1.intersection(set_nums2)
intersect2 = set_nums1.intersection(set_nums3)
... | two-out-of-three | Python fast solution using set, intersection, and union | samanehghafouri | 0 | 23 | two out of three | 2,032 | 0.726 | Easy | 28,275 |
https://leetcode.com/problems/two-out-of-three/discuss/2522189/Simple-python-solution | class Solution:
def twoOutOfThree(self, nums1: List[int], nums2: List[int], nums3: List[int]) -> List[int]:
nums = list(set(nums1))+list(set(nums2))+list(set(nums3))
dic = Counter(nums)
res = []
for k,v in dic.items():
if v >= 2:
res.append(k)
retu... | two-out-of-three | Simple python solution | aruj900 | 0 | 37 | two out of three | 2,032 | 0.726 | Easy | 28,276 |
https://leetcode.com/problems/two-out-of-three/discuss/2418356/Using-Set-Counter-and-list-comprehension-using-Python | class Solution:
def twoOutOfThree(self, nums1: List[int], nums2: List[int], nums3: List[int]) -> List[int]:
n1 = list(set(nums1))
n2 = list(set(nums2))
n3 = list(set(nums3))
n1.extend(n2)
n1.extend(n3)
ct = Counter(n1)
res = [k for k, v in ct.items() if v >= 2... | two-out-of-three | Using Set, Counter and list comprehension using Python | ankurbhambri | 0 | 28 | two out of three | 2,032 | 0.726 | Easy | 28,277 |
https://leetcode.com/problems/two-out-of-three/discuss/2378871/Python-easy-solution-using-List-and-set | class Solution:
def twoOutOfThree(self, nums1: List[int], nums2: List[int], nums3: List[int]) -> List[int]:
ans=[]
s1=list(set(nums1))
s2=list(set(nums2))
s3=list(set(nums3))
for i in s1:
if i in s2 or i in s3:
ans.append(i)
for j in s2:
if j in s1 or j in s3:
ans.append(j)
for k in s3:
... | two-out-of-three | Python easy solution using List and set | keertika27 | 0 | 27 | two out of three | 2,032 | 0.726 | Easy | 28,278 |
https://leetcode.com/problems/two-out-of-three/discuss/2350306/Python-or-Set-or-HashMap | class Solution:
def twoOutOfThree(self, nums1: List[int], nums2: List[int], nums3: List[int]) -> List[int]:
hash_map={}
for i in set(nums1):
hash_map[i]=1+hash_map.get(i,0)
for i in set(nums2):
hash_map[i]=1+hash_map.get(i,0)
for i in set(nums3):
h... | two-out-of-three | Python | Set | HashMap | dinesh1898 | 0 | 20 | two out of three | 2,032 | 0.726 | Easy | 28,279 |
https://leetcode.com/problems/two-out-of-three/discuss/2232602/PYTHON-O(n)-USING-SETS | class Solution:
def twoOutOfThree(self, nums1: List[int], nums2: List[int], nums3: List[int]) -> List[int]:
s1 = set(nums1)
s2 = set(nums2)
s3 = set(nums3)
hash_map = {}
for i in s1:
hash_map[i] = 1
for i in s2:
if i in hash_map:
... | two-out-of-three | PYTHON O(n) USING SETS | akashp2001 | 0 | 49 | two out of three | 2,032 | 0.726 | Easy | 28,280 |
https://leetcode.com/problems/two-out-of-three/discuss/2090093/Python-oneliner | class Solution:
def twoOutOfThree(self, n1: List[int], n2: List[int], n3: List[int]) -> List[int]:
return {item for sublist in [set(n2)&set(n3), set(n1)&set(n2), set(n1)&set(n3)] for item in sublist} | two-out-of-three | Python oneliner | StikS32 | 0 | 65 | two out of three | 2,032 | 0.726 | Easy | 28,281 |
https://leetcode.com/problems/two-out-of-three/discuss/2018781/python3-One-Line-Solution | class Solution:
def twoOutOfThree(self, nums1: List[int], nums2: List[int], nums3: List[int]) -> List[int]:
return (set(nums1) & set(nums2)) | (set(nums1) & set(nums3)) | (set(nums2) & set(nums3)) | two-out-of-three | [python3] One-Line Solution | terrencetang | 0 | 39 | two out of three | 2,032 | 0.726 | Easy | 28,282 |
https://leetcode.com/problems/two-out-of-three/discuss/2001635/Whacky-list-comp-solution | class Solution:
def twoOutOfThree(self, nums1: List[int], nums2: List[int], nums3: List[int]) -> List[int]:
nums = list(set(nums1 + nums2 + nums3))
return [n for n in nums if (n in nums1 and (n in nums2 or n in nums3) or (n in nums2 and n in nums3))] | two-out-of-three | Whacky list comp solution | andrewnerdimo | 0 | 20 | two out of three | 2,032 | 0.726 | Easy | 28,283 |
https://leetcode.com/problems/two-out-of-three/discuss/1938021/easy-python-code | class Solution:
def twoOutOfThree(self, nums1: List[int], nums2: List[int], nums3: List[int]) -> List[int]:
n = list(set(nums1+nums2+nums3))
output = []
for i in n:
if (i in nums1 and (i in nums2 or i in nums3)) or (i in nums2 and i in nums3):
output.append(i)
... | two-out-of-three | easy python code | dakash682 | 0 | 52 | two out of three | 2,032 | 0.726 | Easy | 28,284 |
https://leetcode.com/problems/two-out-of-three/discuss/1937037/Python-Solution-%2B-One-Liner-or-Set-or-Intersection-and-Union | class Solution:
def twoOutOfThree(self, nums1, nums2, nums3):
set1, set2, set3 = set(nums1), set(nums2), set(nums3)
set12 = set1.intersection(set2)
set23 = set2.intersection(set3)
set13 = set1.intersection(set3)
return set12.union(set23).union(set13) | two-out-of-three | Python - Solution + One-Liner | Set | Intersection and Union | domthedeveloper | 0 | 47 | two out of three | 2,032 | 0.726 | Easy | 28,285 |
https://leetcode.com/problems/two-out-of-three/discuss/1937037/Python-Solution-%2B-One-Liner-or-Set-or-Intersection-and-Union | class Solution:
def twoOutOfThree(self, nums1, nums2, nums3):
set1, set2, set3 = set(nums1), set(nums2), set(nums3)
set12 = set1 & set2
set23 = set2 & set3
set13 = set1 & set3
return set12 | set23 | set13 | two-out-of-three | Python - Solution + One-Liner | Set | Intersection and Union | domthedeveloper | 0 | 47 | two out of three | 2,032 | 0.726 | Easy | 28,286 |
https://leetcode.com/problems/two-out-of-three/discuss/1937037/Python-Solution-%2B-One-Liner-or-Set-or-Intersection-and-Union | class Solution:
def twoOutOfThree(self, nums1, nums2, nums3):
set1, set2, set3 = set(nums1), set(nums2), set(nums3)
return set1 & set2 | set2 & set3 | set1 & set3 | two-out-of-three | Python - Solution + One-Liner | Set | Intersection and Union | domthedeveloper | 0 | 47 | two out of three | 2,032 | 0.726 | Easy | 28,287 |
https://leetcode.com/problems/two-out-of-three/discuss/1937037/Python-Solution-%2B-One-Liner-or-Set-or-Intersection-and-Union | class Solution:
def twoOutOfThree(self, nums1, nums2, nums3):
return (lambda s1,s2,s3 : s1&s2 | s2&s3 | s1&s3)(set(nums1),set(nums2),set(nums3)) | two-out-of-three | Python - Solution + One-Liner | Set | Intersection and Union | domthedeveloper | 0 | 47 | two out of three | 2,032 | 0.726 | Easy | 28,288 |
https://leetcode.com/problems/two-out-of-three/discuss/1825044/1-Line-Python-Solution-oror-75-Faster-oror-Memory-less-than-70 | class Solution:
def twoOutOfThree(self, nums1: List[int], nums2: List[int], nums3: List[int]) -> List[int]:
return [i for i,v in Counter([*set(nums1), *set(nums2), *set(nums3)]).most_common() if v>=2] | two-out-of-three | 1-Line Python Solution || 75% Faster || Memory less than 70% | Taha-C | 0 | 70 | two out of three | 2,032 | 0.726 | Easy | 28,289 |
https://leetcode.com/problems/two-out-of-three/discuss/1819465/Python-Easy-to-Understand-Hashmap-Solution-for-beginners | class Solution:
def twoOutOfThree(self, nums1: List[int], nums2: List[int], nums3: List[int]) -> List[int]:
dic1 = {}
dic2 = {}
dic3 = {}
lst = []
for n1 in nums1:
if n1 in dic1:
dic1[n1]+=1
else:
dic1[n1]=1
for ... | two-out-of-three | Python Easy to Understand Hashmap Solution for beginners | Ron99 | 0 | 65 | two out of three | 2,032 | 0.726 | Easy | 28,290 |
https://leetcode.com/problems/two-out-of-three/discuss/1537793/Hashing-%3A-A-beautiful-concept-greater-Thought-Process-Explained | class Solution:
def twoOutOfThree(self, nums1: List[int], nums2: List[int], nums3: List[int]) -> List[int]:
#return a distinct array containing all the values
#that are present in at least two out of the three arrays.
#NOTE - values can be in any order
d1 ... | two-out-of-three | Hashing : A beautiful concept --> Thought Process Explained | aarushsharmaa | 0 | 114 | two out of three | 2,032 | 0.726 | Easy | 28,291 |
https://leetcode.com/problems/two-out-of-three/discuss/1528452/Python3-Two-kind-of-solutions | class Solution:
def twoOutOfThree(self, nums1: List[int], nums2: List[int], nums3: List[int]) -> List[int]:
nums1 = set(nums1)
nums2 = set(nums2)
nums3 = set(nums3)
d = {}
ans = []
for num in nums1:
if num not in d:
d[num]... | two-out-of-three | [Python3] Two kind of solutions | maosipov11 | 0 | 73 | two out of three | 2,032 | 0.726 | Easy | 28,292 |
https://leetcode.com/problems/two-out-of-three/discuss/1528452/Python3-Two-kind-of-solutions | class Solution:
def twoOutOfThree(self, nums1: List[int], nums2: List[int], nums3: List[int]) -> List[int]:
nums1 = set(nums1)
nums2 = set(nums2)
nums3 = set(nums3)
return list((nums1&nums2) | (nums2&nums3) | (nums1&nums3)) | two-out-of-three | [Python3] Two kind of solutions | maosipov11 | 0 | 73 | two out of three | 2,032 | 0.726 | Easy | 28,293 |
https://leetcode.com/problems/two-out-of-three/discuss/1523046/Python-or-list-or-set | class Solution:
def twoOutOfThree(self, nums1: List[int], nums2: List[int], nums3: List[int]) -> List[int]:
out = []
for key in set(nums1):
if key in set(nums2) or key in set(nums3):
out.append(key)
for key in set(nums2):
if key in set... | two-out-of-three | Python | list | set | _rust | 0 | 69 | two out of three | 2,032 | 0.726 | Easy | 28,294 |
https://leetcode.com/problems/two-out-of-three/discuss/1520972/Python-Solution-or-Using-Dictionary | class Solution:
def twoOutOfThree(self, nums1: List[int], nums2: List[int], nums3: List[int]) -> List[int]:
a = []
table = defaultdict(int)
# to remove duplicates from each list
first = list(OrderedDict.fromkeys(nums1))
second = list(OrderedDict.fromkeys(nums2))
... | two-out-of-three | Python Solution | Using Dictionary | Shreya19595 | 0 | 101 | two out of three | 2,032 | 0.726 | Easy | 28,295 |
https://leetcode.com/problems/two-out-of-three/discuss/1514408/Python-one-liner | class Solution:
def twoOutOfThree(self, nums1: List[int], nums2: List[int], nums3: List[int]) -> List[int]:
return [v for v, c in Counter(itertools.chain(set(nums1), set(nums2), set(nums3))).items() if c >=2] | two-out-of-three | Python, one-liner | blue_sky5 | 0 | 64 | two out of three | 2,032 | 0.726 | Easy | 28,296 |
https://leetcode.com/problems/minimum-operations-to-make-a-uni-value-grid/discuss/1513319/Python3-median-4-line | class Solution:
def minOperations(self, grid: List[List[int]], x: int) -> int:
vals = [x for row in grid for x in row]
if len(set(val%x for val in vals)) > 1: return -1 # impossible
median = sorted(vals)[len(vals)//2] # O(N) possible via "quick select"
return sum(abs(val - median)//x... | minimum-operations-to-make-a-uni-value-grid | [Python3] median 4-line | ye15 | 21 | 1,100 | minimum operations to make a uni value grid | 2,033 | 0.524 | Medium | 28,297 |
https://leetcode.com/problems/minimum-operations-to-make-a-uni-value-grid/discuss/1785741/Python3-Why-median-is-optimal-or-Detailed-Explanation | class Solution:
def minOperations(self, grid: List[List[int]], x: int) -> int:
li = []
# convert matrix to array, we dont care about the structure itself. We just want the values
for val in grid:
li+= val
# sort the array
li.sort()
... | minimum-operations-to-make-a-uni-value-grid | [Python3] Why median is optimal | Detailed Explanation | Apurv_Moroney | 4 | 278 | minimum operations to make a uni value grid | 2,033 | 0.524 | Medium | 28,298 |
https://leetcode.com/problems/minimum-operations-to-make-a-uni-value-grid/discuss/1513305/Median-or-Python-or-Simple-Approach-or-Explanation-with-Comments | class Solution:
def minOperations(self, grid: List[List[int]], x: int) -> int:
# flatten the numbers
nums = []
for row in grid:
for num in row:
nums.append(num)
# sort and find the median
nums.sort()
n = len(nums)
... | minimum-operations-to-make-a-uni-value-grid | Median | Python | Simple Approach | Explanation with Comments | CaptainX | 2 | 163 | minimum operations to make a uni value grid | 2,033 | 0.524 | Medium | 28,299 |
Subsets and Splits
Top 2 Solutions by Upvotes
Identifies the top 2 highest upvoted Python solutions for each problem, providing insight into popular approaches.